I'm taking my first course of Analysis and read about this Theorem (2.35) in Rudin:
Closed subsets of compact sets are compact.
I want to know that whether open subsets of compact sets are compact? But I was looking at the proof, which uses the knowledge of $F \subset X \subset X$ and $F$ is closed relative to $X$ and $K$ is compact. But I have no idea how to approach the proof with open $F$. Thank you!
I was then thinking about the preceding Theorem (2.34) which says:
Compact subsets of metric spaces are closed.
Can we use the converse of this to directly disprove the statement? Since for $K \subset Y \subset X$ where $Y$ is compact subset of $X$, $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$. So by Theorem 2.34, an open subset $K$ are not compact relative to $X$, thus $K$ is not compact relative to $Y$ as well. Does this seem reasonable?
Best Answer
I just want to explain why the example, that everybody mentioned, $(0,1)\subset [0,1]$ is indeed not compact. I'm doing it since although this is really easy to see using the sequential compact definition, using the cover definition of compactness it is not as trivial to see (well not to the eyes of a first timer).
By sequential definition of compactness: Take sequence $a_n = 1/n$ which converges to $0$ which is outside of $(0,1)$ although $a_n\in (0,1)$ for all $n$. Furthermore no subsequence of $a_n$ converges in $(0,1)$. So $(0,1)$ is not compact.
Using covers: Let's engineer an infinite set of open sets covering $(0,1)$ with no finite sub-cover. Take $U_1=(\frac{1}{2},1)$ and $U_n = (\frac{1}{2^{n}},\frac{3}{2^{n}})$. This is a covering of $(0,1)$. But then $\frac{1}{2^{m}}$ is only covered by $U_{m+1}$ and no other open set in the cover. Take any finite sub-collection of these open sets, then you are missing a lot of $\frac{1}{2^m}$ not covered by anything. So there is no finite subcover, and no compactness.